/src/openssl/crypto/bn/bn_sqrt.c

Source (jump to first uncovered line)
/*
 * Copyright 2000-2018 The OpenSSL Project Authors. All Rights Reserved.
 *
 * Licensed under the OpenSSL license (the "License").  You may not use
 * this file except in compliance with the License.  You can obtain a copy
 * in the file LICENSE in the source distribution or at
 * https://www.openssl.org/source/license.html
 */

#include "internal/cryptlib.h"
#include "bn_lcl.h"

BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
/*
 * Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
 * algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
 * Theory", algorithm 1.5.1). 'p' must be prime!
 */
{
    BIGNUM *ret = in;
    int err = 1;
    int r;
    BIGNUM *A, *b, *q, *t, *x, *y;
    int e, i, j;

    if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
        if (BN_abs_is_word(p, 2)) {
            if (ret == NULL)
                ret = BN_new();
            if (ret == NULL)
                goto end;
            if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
                if (ret != in)
                    BN_free(ret);
                return NULL;
            }
            bn_check_top(ret);
            return ret;
        }

        BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
        return NULL;
    }

    if (BN_is_zero(a) || BN_is_one(a)) {
        if (ret == NULL)
            ret = BN_new();
        if (ret == NULL)
            goto end;
        if (!BN_set_word(ret, BN_is_one(a))) {
            if (ret != in)
                BN_free(ret);
            return NULL;
        }
        bn_check_top(ret);
        return ret;
    }

    BN_CTX_start(ctx);
    A = BN_CTX_get(ctx);
    b = BN_CTX_get(ctx);
    q = BN_CTX_get(ctx);
    t = BN_CTX_get(ctx);
    x = BN_CTX_get(ctx);
    y = BN_CTX_get(ctx);
    if (y == NULL)
        goto end;

    if (ret == NULL)
        ret = BN_new();
    if (ret == NULL)
        goto end;

    /* A = a mod p */
    if (!BN_nnmod(A, a, p, ctx))
        goto end;

    /* now write  |p| - 1  as  2^e*q  where  q  is odd */
    e = 1;
    while (!BN_is_bit_set(p, e))
        e++;
    /* we'll set  q  later (if needed) */

    if (e == 1) {
        /*-
         * The easy case:  (|p|-1)/2  is odd, so 2 has an inverse
         * modulo  (|p|-1)/2,  and square roots can be computed
         * directly by modular exponentiation.
         * We have
         *     2 * (|p|+1)/4 == 1   (mod (|p|-1)/2),
         * so we can use exponent  (|p|+1)/4,  i.e.  (|p|-3)/4 + 1.
         */
        if (!BN_rshift(q, p, 2))
            goto end;
        q->neg = 0;
        if (!BN_add_word(q, 1))
            goto end;
        if (!BN_mod_exp(ret, A, q, p, ctx))
            goto end;
        err = 0;
        goto vrfy;
    }

    if (e == 2) {
        /*-
         * |p| == 5  (mod 8)
         *
         * In this case  2  is always a non-square since
         * Legendre(2,p) = (-1)^((p^2-1)/8)  for any odd prime.
         * So if  a  really is a square, then  2*a  is a non-square.
         * Thus for
         *      b := (2*a)^((|p|-5)/8),
         *      i := (2*a)*b^2
         * we have
         *     i^2 = (2*a)^((1 + (|p|-5)/4)*2)
         *         = (2*a)^((p-1)/2)
         *         = -1;
         * so if we set
         *      x := a*b*(i-1),
         * then
         *     x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
         *         = a^2 * b^2 * (-2*i)
         *         = a*(-i)*(2*a*b^2)
         *         = a*(-i)*i
         *         = a.
         *
         * (This is due to A.O.L. Atkin,
         * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
         * November 1992.)
         */

        /* t := 2*a */
        if (!BN_mod_lshift1_quick(t, A, p))
            goto end;

        /* b := (2*a)^((|p|-5)/8) */
        if (!BN_rshift(q, p, 3))
            goto end;
        q->neg = 0;
        if (!BN_mod_exp(b, t, q, p, ctx))
            goto end;

        /* y := b^2 */
        if (!BN_mod_sqr(y, b, p, ctx))
            goto end;

        /* t := (2*a)*b^2 - 1 */
        if (!BN_mod_mul(t, t, y, p, ctx))
            goto end;
        if (!BN_sub_word(t, 1))
            goto end;

        /* x = a*b*t */
        if (!BN_mod_mul(x, A, b, p, ctx))
            goto end;
        if (!BN_mod_mul(x, x, t, p, ctx))
            goto end;

        if (!BN_copy(ret, x))
            goto end;
        err = 0;
        goto vrfy;
    }

    /*
     * e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
     * find some y that is not a square.
     */
    if (!BN_copy(q, p))
        goto end;               /* use 'q' as temp */
    q->neg = 0;
    i = 2;
    do {
        /*
         * For efficiency, try small numbers first; if this fails, try random
         * numbers.
         */
        if (i < 22) {
            if (!BN_set_word(y, i))
                goto end;
        } else {
            if (!BN_priv_rand(y, BN_num_bits(p), 0, 0))
                goto end;
            if (BN_ucmp(y, p) >= 0) {
                if (!(p->neg ? BN_add : BN_sub) (y, y, p))
                    goto end;
            }
            /* now 0 <= y < |p| */
            if (BN_is_zero(y))
                if (!BN_set_word(y, i))
                    goto end;
        }

        r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
        if (r < -1)
            goto end;
        if (r == 0) {
            /* m divides p */
            BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
            goto end;
        }
    }
    while (r == 1 && ++i < 82);

    if (r != -1) {
        /*
         * Many rounds and still no non-square -- this is more likely a bug
         * than just bad luck. Even if p is not prime, we should have found
         * some y such that r == -1.
         */
        BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
        goto end;
    }

    /* Here's our actual 'q': */
    if (!BN_rshift(q, q, e))
        goto end;

    /*
     * Now that we have some non-square, we can find an element of order 2^e
     * by computing its q'th power.
     */
    if (!BN_mod_exp(y, y, q, p, ctx))
        goto end;
    if (BN_is_one(y)) {
        BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
        goto end;
    }

    /*-
     * Now we know that (if  p  is indeed prime) there is an integer
     * k,  0 <= k < 2^e,  such that
     *
     *      a^q * y^k == 1   (mod p).
     *
     * As  a^q  is a square and  y  is not,  k  must be even.
     * q+1  is even, too, so there is an element
     *
     *     X := a^((q+1)/2) * y^(k/2),
     *
     * and it satisfies
     *
     *     X^2 = a^q * a     * y^k
     *         = a,
     *
     * so it is the square root that we are looking for.
     */

    /* t := (q-1)/2  (note that  q  is odd) */
    if (!BN_rshift1(t, q))
        goto end;

    /* x := a^((q-1)/2) */
    if (BN_is_zero(t)) {        /* special case: p = 2^e + 1 */
        if (!BN_nnmod(t, A, p, ctx))
            goto end;
        if (BN_is_zero(t)) {
            /* special case: a == 0  (mod p) */
            BN_zero(ret);
            err = 0;
            goto end;
        } else if (!BN_one(x))
            goto end;
    } else {
        if (!BN_mod_exp(x, A, t, p, ctx))
            goto end;
        if (BN_is_zero(x)) {
            /* special case: a == 0  (mod p) */
            BN_zero(ret);
            err = 0;
            goto end;
        }
    }

    /* b := a*x^2  (= a^q) */
    if (!BN_mod_sqr(b, x, p, ctx))
        goto end;
    if (!BN_mod_mul(b, b, A, p, ctx))
        goto end;

    /* x := a*x    (= a^((q+1)/2)) */
    if (!BN_mod_mul(x, x, A, p, ctx))
        goto end;

    while (1) {
        /*-
         * Now  b  is  a^q * y^k  for some even  k  (0 <= k < 2^E
         * where  E  refers to the original value of  e,  which we
         * don't keep in a variable),  and  x  is  a^((q+1)/2) * y^(k/2).
         *
         * We have  a*b = x^2,
         *    y^2^(e-1) = -1,
         *    b^2^(e-1) = 1.
         */

        if (BN_is_one(b)) {
            if (!BN_copy(ret, x))
                goto end;
            err = 0;
            goto vrfy;
        }

        /* find smallest  i  such that  b^(2^i) = 1 */
        i = 1;
        if (!BN_mod_sqr(t, b, p, ctx))
            goto end;
        while (!BN_is_one(t)) {
            i++;
            if (i == e) {
                BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
                goto end;
            }
            if (!BN_mod_mul(t, t, t, p, ctx))
                goto end;
        }

        /* t := y^2^(e - i - 1) */
        if (!BN_copy(t, y))
            goto end;
        for (j = e - i - 1; j > 0; j--) {
            if (!BN_mod_sqr(t, t, p, ctx))
                goto end;
        }
        if (!BN_mod_mul(y, t, t, p, ctx))
            goto end;
        if (!BN_mod_mul(x, x, t, p, ctx))
            goto end;
        if (!BN_mod_mul(b, b, y, p, ctx))
            goto end;
        e = i;
    }

 vrfy:
    if (!err) {
        /*
         * verify the result -- the input might have been not a square (test
         * added in 0.9.8)
         */

        if (!BN_mod_sqr(x, ret, p, ctx))
            err = 1;

        if (!err && 0 != BN_cmp(x, A)) {
            BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
            err = 1;
        }
    }

 end:
    if (err) {
        if (ret != in)
            BN_clear_free(ret);
        ret = NULL;
    }
    BN_CTX_end(ctx);
    bn_check_top(ret);
    return ret;
}

Coverage Report

Created: 2018-08-29 13:53

Line	Count	Source (jump to first uncovered line)
1		/*
2		* Copyright 2000-2018 The OpenSSL Project Authors. All Rights Reserved.
3		*
4		* Licensed under the OpenSSL license (the "License"). You may not use
5		* this file except in compliance with the License. You can obtain a copy
6		* in the file LICENSE in the source distribution or at
7		* https://www.openssl.org/source/license.html
8		*/
9
10		#include "internal/cryptlib.h"
11		#include "bn_lcl.h"
12
13		BIGNUM BN_mod_sqrt(BIGNUM in, const BIGNUM a, const BIGNUM p, BN_CTX *ctx)
14		/*
15		* Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
16		* algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
17		* Theory", algorithm 1.5.1). 'p' must be prime!
18		*/
19	0	{
20	0	BIGNUM *ret = in;
21	0	int err = 1;
22	0	int r;
23	0	BIGNUM A, b, q, t, x, y;
24	0	int e, i, j;
25	0
26	0	if (!BN_is_odd(p) \|\| BN_abs_is_word(p, 1)) {
27	0	if (BN_abs_is_word(p, 2)) {
28	0	if (ret == NULL)
29	0	ret = BN_new();
30	0	if (ret == NULL)
31	0	goto end;
32	0	if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
33	0	if (ret != in)
34	0	BN_free(ret);
35	0	return NULL;
36	0	}
37	0	bn_check_top(ret);
38	0	return ret;
39	0	}
40	0
41	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
42	0	return NULL;
43	0	}
44	0
45	0	if (BN_is_zero(a) \|\| BN_is_one(a)) {
46	0	if (ret == NULL)
47	0	ret = BN_new();
48	0	if (ret == NULL)
49	0	goto end;
50	0	if (!BN_set_word(ret, BN_is_one(a))) {
51	0	if (ret != in)
52	0	BN_free(ret);
53	0	return NULL;
54	0	}
55	0	bn_check_top(ret);
56	0	return ret;
57	0	}
58	0
59	0	BN_CTX_start(ctx);
60	0	A = BN_CTX_get(ctx);
61	0	b = BN_CTX_get(ctx);
62	0	q = BN_CTX_get(ctx);
63	0	t = BN_CTX_get(ctx);
64	0	x = BN_CTX_get(ctx);
65	0	y = BN_CTX_get(ctx);
66	0	if (y == NULL)
67	0	goto end;
68	0
69	0	if (ret == NULL)
70	0	ret = BN_new();
71	0	if (ret == NULL)
72	0	goto end;
73	0
74	0	/* A = a mod p */
75	0	if (!BN_nnmod(A, a, p, ctx))
76	0	goto end;
77	0
78	0	/* now write \|p\| - 1 as 2^eq where q is odd /
79	0	e = 1;
80	0	while (!BN_is_bit_set(p, e))
81	0	e++;
82	0	/* we'll set q later (if needed) */
83	0
84	0	if (e == 1) {
85	0	/*-
86	0	* The easy case: (\|p\|-1)/2 is odd, so 2 has an inverse
87	0	* modulo (\|p\|-1)/2, and square roots can be computed
88	0	* directly by modular exponentiation.
89	0	* We have
90	0	* 2 * (\|p\|+1)/4 == 1 (mod (\|p\|-1)/2),
91	0	* so we can use exponent (\|p\|+1)/4, i.e. (\|p\|-3)/4 + 1.
92	0	*/
93	0	if (!BN_rshift(q, p, 2))
94	0	goto end;
95	0	q->neg = 0;
96	0	if (!BN_add_word(q, 1))
97	0	goto end;
98	0	if (!BN_mod_exp(ret, A, q, p, ctx))
99	0	goto end;
100	0	err = 0;
101	0	goto vrfy;
102	0	}
103	0
104	0	if (e == 2) {
105	0	/*-
106	0	* \|p\| == 5 (mod 8)
107	0	*
108	0	* In this case 2 is always a non-square since
109	0	* Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
110	0	* So if a really is a square, then 2*a is a non-square.
111	0	* Thus for
112	0	* b := (2*a)^((\|p\|-5)/8),
113	0	* i := (2a)b^2
114	0	* we have
115	0	* i^2 = (2a)^((1 + (\|p\|-5)/4)2)
116	0	* = (2*a)^((p-1)/2)
117	0	* = -1;
118	0	* so if we set
119	0	* x := ab(i-1),
120	0	* then
121	0	* x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
122	0	* = a^2 * b^2 * (-2*i)
123	0	* = a(-i)(2ab^2)
124	0	* = a(-i)i
125	0	* = a.
126	0	*
127	0	* (This is due to A.O.L. Atkin,
128	0	* <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
129	0	* November 1992.)
130	0	*/
131	0
132	0	/* t := 2a /
133	0	if (!BN_mod_lshift1_quick(t, A, p))
134	0	goto end;
135	0
136	0	/* b := (2a)^((\|p\|-5)/8) /
137	0	if (!BN_rshift(q, p, 3))
138	0	goto end;
139	0	q->neg = 0;
140	0	if (!BN_mod_exp(b, t, q, p, ctx))
141	0	goto end;
142	0
143	0	/* y := b^2 */
144	0	if (!BN_mod_sqr(y, b, p, ctx))
145	0	goto end;
146	0
147	0	/* t := (2a)b^2 - 1 */
148	0	if (!BN_mod_mul(t, t, y, p, ctx))
149	0	goto end;
150	0	if (!BN_sub_word(t, 1))
151	0	goto end;
152	0
153	0	/* x = abt */
154	0	if (!BN_mod_mul(x, A, b, p, ctx))
155	0	goto end;
156	0	if (!BN_mod_mul(x, x, t, p, ctx))
157	0	goto end;
158	0
159	0	if (!BN_copy(ret, x))
160	0	goto end;
161	0	err = 0;
162	0	goto vrfy;
163	0	}
164	0
165	0	/*
166	0	* e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
167	0	* find some y that is not a square.
168	0	*/
169	0	if (!BN_copy(q, p))
170	0	goto end; /* use 'q' as temp */
171	0	q->neg = 0;
172	0	i = 2;
173	0	do {
174	0	/*
175	0	* For efficiency, try small numbers first; if this fails, try random
176	0	* numbers.
177	0	*/
178	0	if (i < 22) {
179	0	if (!BN_set_word(y, i))
180	0	goto end;
181	0	} else {
182	0	if (!BN_priv_rand(y, BN_num_bits(p), 0, 0))
183	0	goto end;
184	0	if (BN_ucmp(y, p) >= 0) {
185	0	if (!(p->neg ? BN_add : BN_sub) (y, y, p))
186	0	goto end;
187	0	}
188	0	/* now 0 <= y < \|p\| */
189	0	if (BN_is_zero(y))
190	0	if (!BN_set_word(y, i))
191	0	goto end;
192	0	}
193	0
194	0	r = BN_kronecker(y, q, ctx); /* here 'q' is \|p\| */
195	0	if (r < -1)
196	0	goto end;
197	0	if (r == 0) {
198	0	/* m divides p */
199	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
200	0	goto end;
201	0	}
202	0	}
203	0	while (r == 1 && ++i < 82);
204	0
205	0	if (r != -1) {
206	0	/*
207	0	* Many rounds and still no non-square -- this is more likely a bug
208	0	* than just bad luck. Even if p is not prime, we should have found
209	0	* some y such that r == -1.
210	0	*/
211	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
212	0	goto end;
213	0	}
214	0
215	0	/* Here's our actual 'q': */
216	0	if (!BN_rshift(q, q, e))
217	0	goto end;
218	0
219	0	/*
220	0	* Now that we have some non-square, we can find an element of order 2^e
221	0	* by computing its q'th power.
222	0	*/
223	0	if (!BN_mod_exp(y, y, q, p, ctx))
224	0	goto end;
225	0	if (BN_is_one(y)) {
226	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
227	0	goto end;
228	0	}
229	0
230	0	/*-
231	0	* Now we know that (if p is indeed prime) there is an integer
232	0	* k, 0 <= k < 2^e, such that
233	0	*
234	0	* a^q * y^k == 1 (mod p).
235	0	*
236	0	* As a^q is a square and y is not, k must be even.
237	0	* q+1 is even, too, so there is an element
238	0	*
239	0	* X := a^((q+1)/2) * y^(k/2),
240	0	*
241	0	* and it satisfies
242	0	*
243	0	* X^2 = a^q * a * y^k
244	0	* = a,
245	0	*
246	0	* so it is the square root that we are looking for.
247	0	*/
248	0
249	0	/* t := (q-1)/2 (note that q is odd) */
250	0	if (!BN_rshift1(t, q))
251	0	goto end;
252	0
253	0	/* x := a^((q-1)/2) */
254	0	if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
255	0	if (!BN_nnmod(t, A, p, ctx))
256	0	goto end;
257	0	if (BN_is_zero(t)) {
258	0	/* special case: a == 0 (mod p) */
259	0	BN_zero(ret);
260	0	err = 0;
261	0	goto end;
262	0	} else if (!BN_one(x))
263	0	goto end;
264	0	} else {
265	0	if (!BN_mod_exp(x, A, t, p, ctx))
266	0	goto end;
267	0	if (BN_is_zero(x)) {
268	0	/* special case: a == 0 (mod p) */
269	0	BN_zero(ret);
270	0	err = 0;
271	0	goto end;
272	0	}
273	0	}
274	0
275	0	/* b := ax^2 (= a^q) /
276	0	if (!BN_mod_sqr(b, x, p, ctx))
277	0	goto end;
278	0	if (!BN_mod_mul(b, b, A, p, ctx))
279	0	goto end;
280	0
281	0	/* x := ax (= a^((q+1)/2)) /
282	0	if (!BN_mod_mul(x, x, A, p, ctx))
283	0	goto end;
284	0
285	0	while (1) {
286	0	/*-
287	0	* Now b is a^q * y^k for some even k (0 <= k < 2^E
288	0	* where E refers to the original value of e, which we
289	0	* don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
290	0	*
291	0	* We have a*b = x^2,
292	0	* y^2^(e-1) = -1,
293	0	* b^2^(e-1) = 1.
294	0	*/
295	0
296	0	if (BN_is_one(b)) {
297	0	if (!BN_copy(ret, x))
298	0	goto end;
299	0	err = 0;
300	0	goto vrfy;
301	0	}
302	0
303	0	/* find smallest i such that b^(2^i) = 1 */
304	0	i = 1;
305	0	if (!BN_mod_sqr(t, b, p, ctx))
306	0	goto end;
307	0	while (!BN_is_one(t)) {
308	0	i++;
309	0	if (i == e) {
310	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
311	0	goto end;
312	0	}
313	0	if (!BN_mod_mul(t, t, t, p, ctx))
314	0	goto end;
315	0	}
316	0
317	0	/* t := y^2^(e - i - 1) */
318	0	if (!BN_copy(t, y))
319	0	goto end;
320	0	for (j = e - i - 1; j > 0; j--) {
321	0	if (!BN_mod_sqr(t, t, p, ctx))
322	0	goto end;
323	0	}
324	0	if (!BN_mod_mul(y, t, t, p, ctx))
325	0	goto end;
326	0	if (!BN_mod_mul(x, x, t, p, ctx))
327	0	goto end;
328	0	if (!BN_mod_mul(b, b, y, p, ctx))
329	0	goto end;
330	0	e = i;
331	0	}
332	0
333	0	vrfy:
334	0	if (!err) {
335	0	/*
336	0	* verify the result -- the input might have been not a square (test
337	0	* added in 0.9.8)
338	0	*/
339	0
340	0	if (!BN_mod_sqr(x, ret, p, ctx))
341	0	err = 1;
342	0
343	0	if (!err && 0 != BN_cmp(x, A)) {
344	0	BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
345	0	err = 1;
346	0	}
347	0	}
348	0
349	0	end:
350	0	if (err) {
351	0	if (ret != in)
352	0	BN_clear_free(ret);
353	0	ret = NULL;
354	0	}
355	0	BN_CTX_end(ctx);
356	0	bn_check_top(ret);
357	0	return ret;
358	0	}